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About LinkBacksThe given relation S on real numbers, R, is given by
xSy <=> x*y > 0
Determine wether S is reflexive, symmetric and transitive.
How can I solve this assignment by proof or contradiction?
Note that reflexivity fails. Can you find the single element which this does not hold for? This is your contradiction.Originally Posted by Madspeter
However, the other two do hold. Symmetry is pretty simple (it comes from the commutativity of * over). For transitivity, we can use the fact that
. Two cases:
(1)
.
(2) is identical, but we take.
The element which you speak of must be the fact that either x or y can be chosen as 0 - giving us a false statement. 0 is not bigger than 0.
Is this how you mean reflexitvity fails?
suppose in your 1) that x = 0 instead of x < 0 - wouldnt this give us a non-transitive relation?
of courseThat is correct.
Rethink the transitive part. Ifare related that is saying that they are not zero and both have the same sign.
if xSy <=> x*y > 0
then either x or y can't be 0, cause they have to > 0, because they are related.
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this, however, does not hold for the reflexivity, where x can be 0.
right?
Do you think you have timefor one more task of relation? if not, have a nice evening and of course thanks for your help with xSy, it brought a bit more insight to relations.
Here goes.
The relation T on positive R+ = {x E R | x > 0 } is given by:
xTy <=> [x E Q and x*y E Q]
Determine reflexivity, symmetry and transivity. Use proof and contradiction.
Im not sure wether to be confused by the extra amount of structere in this relation or to recognize this as very similar to the prior assignment.
I'd go for not reflexive since either x or y can be 0
The relation symmetric (commutitave rule, (x*y)=(y*x)
though for transitive relation, we can't use your 1) and have to proof xRz by having x>0
am I far off?
No - the relation is both symmetric and transitive because 0 is related to nothing. If we havethen neither
nor
is 0 (as
).
As for your other problem, you should note that for, and also that
. You can use this result to show that your relation is, in fact, an equivalence relation.
That should read "is both symmetric and transitive", not reflexive. I've edited my previous post. I hope that stops any confusion...?I need to clarify a couple of things.
quote
No - the relation is both reflexive and transitive because 0 is related to nothing. If we havethen neither
nor
is 0 (as
).
quote
I thought you meant xSy = x*y > 0 failed to be reflexive?
this is very clear!
as goes for relation T:
quote
The relation T on positive R+ = {x E R | x > 0 } is given by:
xTy <=> [x E Q and x*y E Q]
quote
Since x has to be bigger than 0, then my guess is that T is reflexive.
x is related to x since any x*x > 0.
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, thanks.
